How can I prove if $x>0$ then $-x<0$?
$$x>0implies-x<0$$
I thought about using the axioms of multiplication with $xcdot (-1) = -x$
but I am kind of stuck there.
inequality arithmetic
edited Nov 25 at 14:07
amWhy
191k28224439
asked Nov 25 at 14:00
Another Noone
32
add a comment |
$$x>0implies-x<0$$
I thought about using the axioms of multiplication with $xcdot (-1) = -x$
but I am kind of stuck there.
inequality arithmetic
edited Nov 25 at 14:07
amWhy
191k28224439
asked Nov 25 at 14:00
Another Noone
32
9
How about adding $-x$ to both sides?
– Lord Shark the Unknown
Nov 25 at 14:00
I feel stupid. But thank you very much!
– Another Noone
Nov 25 at 14:02
add a comment |
$$x>0implies-x<0$$
I thought about using the axioms of multiplication with $xcdot (-1) = -x$
but I am kind of stuck there.
inequality arithmetic
edited Nov 25 at 14:07
amWhy
191k28224439
asked Nov 25 at 14:00
Another Noone
32
$$x>0implies-x<0$$
I thought about using the axioms of multiplication with $xcdot (-1) = -x$
but I am kind of stuck there.
inequality arithmetic
inequality arithmetic
edited Nov 25 at 14:07
amWhy
191k28224439
asked Nov 25 at 14:00
Another Noone
32
edited Nov 25 at 14:07
amWhy
191k28224439
asked Nov 25 at 14:00
Another Noone
32
edited Nov 25 at 14:07
amWhy
191k28224439
edited Nov 25 at 14:07
amWhy
191k28224439
edited Nov 25 at 14:07
amWhy
191k28224439
191k28224439
asked Nov 25 at 14:00
Another Noone
32
asked Nov 25 at 14:00
Another Noone
32
asked Nov 25 at 14:00
Another Noone
32
32
9
How about adding $-x$ to both sides?
– Lord Shark the Unknown
Nov 25 at 14:00
I feel stupid. But thank you very much!
– Another Noone
Nov 25 at 14:02
add a comment |
9
How about adding $-x$ to both sides?
– Lord Shark the Unknown
Nov 25 at 14:00
I feel stupid. But thank you very much!
– Another Noone
Nov 25 at 14:02
9
9
How about adding $-x$ to both sides?
– Lord Shark the Unknown
Nov 25 at 14:00
How about adding $-x$ to both sides?
– Lord Shark the Unknown
Nov 25 at 14:00
I feel stupid. But thank you very much!
– Another Noone
Nov 25 at 14:02
I feel stupid. But thank you very much!
– Another Noone
Nov 25 at 14:02
add a comment |
1 Answer
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active
oldest
votes
If x>0 then 0=-x+x>-x+0 so that -x<0. If x<0 then 0=-x+x<-x+0 so that -x>0
answered Nov 25 at 14:47
John Nash
6818
add a comment |
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If x>0 then 0=-x+x>-x+0 so that -x<0. If x<0 then 0=-x+x<-x+0 so that -x>0
answered Nov 25 at 14:47
John Nash
6818
add a comment |
If x>0 then 0=-x+x>-x+0 so that -x<0. If x<0 then 0=-x+x<-x+0 so that -x>0
answered Nov 25 at 14:47
John Nash
6818
add a comment |
If x>0 then 0=-x+x>-x+0 so that -x<0. If x<0 then 0=-x+x<-x+0 so that -x>0
answered Nov 25 at 14:47
John Nash
6818
If x>0 then 0=-x+x>-x+0 so that -x<0. If x<0 then 0=-x+x<-x+0 so that -x>0
answered Nov 25 at 14:47
John Nash
6818
answered Nov 25 at 14:47
John Nash
6818
answered Nov 25 at 14:47
John Nash
6818
answered Nov 25 at 14:47
John Nash
6818
6818
add a comment |
add a comment |
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9
How about adding $-x$ to both sides?
– Lord Shark the Unknown
Nov 25 at 14:00
I feel stupid. But thank you very much!
– Another Noone
Nov 25 at 14:02